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Maths RMO Problems(1)........

Hey friends.

I mean Brilliantians I am back with some amazing problems which are generally asked in the RMO-INMO level examination .I am sharing the image of the paper containing the questions.

Please try and if possible send the solutions

.Also it would be great if you all participate in sharing the questions from your own. I would also be sharing problems based on NSEP level. Thanks

Dont forget to share and like this.............

Note by Abhisek Mohanty
1 month, 2 weeks ago

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Question 2

\[(xy-7)^2=x^2+y^2\Rightarrow x^2y^2-14xy+49=x^2+y^2 \\ x^2y^2-12xy+36+13=x^2+y^2+2xy \\ (x+y+xy-6)(x+y-xy+6)=13=13×1=1×13=-13×-1=-1×-13 \\ (x,y)=(3,4),(4,3),(0,7),(7,0)\] Akshat Sharda · 1 month, 2 weeks ago

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@Akshat Sharda Nice solution................upvoted.... Abhisek Mohanty · 1 month, 2 weeks ago

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question 11 given 34x=43y =>34x+43x=43(y+x) =>77x=43(x+y) now 43 does not divide 77 hence x+y contains 77 i.e-11*7 hence x+y is not prime. Alekhya China · 1 month, 1 week ago

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@Alekhya China That does not exclude the possibility of x+y being odd but not prime Abdur Rehman Zahid · 1 month, 1 week ago

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@Abdur Rehman Zahid Can you please explain me what u are trying to say? Alekhya China · 1 month, 1 week ago

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Which grade you in ? Rajdeep Dhingra · 1 month, 1 week ago

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Question 9 We shall analyze 2 cases. Case 1 Either of \(a,b\) is even.

WLOG let \(a\) be even.Then \(ab(a-b)=45045\) is even.But 45045 is odd,contradiction. Hence no solutions exist in this case.

Case 2 Both \(a,b\) are odd.

Then \(a-b\) is even.Therefore \(ab(a-b)=45045\) is even,contradiction.

Hence no solutions exist. Abdur Rehman Zahid · 1 month, 1 week ago

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Question 8(i)

Let the roots be \(a,b,c,d\).(Note that the roots are positive).Then: \[\begin{align} p&=-(a+b+c+d)\\ q&=ab+ac+ad+bc+bd+cd\\ r&=-(abc+abd+acd+bcd)\\ s&=abcd \end{align}\] \(pr-16s\geq 0\implies (a+b+c+d)(abc+abd+acd+bcd)\geq 16abcd\) which follows by applying AM-GM on each term.

I couldn't understand;what does the variable "a" denote in Q 8(ii)? Abdur Rehman Zahid · 1 month, 2 weeks ago

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Question 1 \[\begin{align} x &\equiv 0,1,2,3,4,5,6 \pmod{7}\\ x^3 &\equiv 0,1,-1\pmod{7} \end{align}\]

Assume that either of \(a,b,c\) is a multiple of 7.

Then obviously \(abc(a^3-b^3)(b^3-c^3)(c^3-a^3)\) is a multiple of 7.

So now WLOG assume that \(a,b,c\not\equiv 0\pmod{7}\).Then \(a^3,b^3,c^3\equiv 1,-1\pmod{7}\).There are \(2\times 2\times 2=8\) different possible cases corresponding to the different values of \(a^3,b^3\) and \(c^3\) modulo 7,which are: \[\begin{array}{c|c|c|c} \text{Values modulo 7} & a^3 & b^3 & c^3 \\ \hline \text{Case 1} & 1 & 1 & 1 \\ \hline \text{Case 2} & -1 & -1 & -1 \\ \hline \text{Case 3} & 1 & -1 & -1 \\ \hline \text{Case 4} & -1 & 1 & -1 \\ \hline \text{Case 5} & -1 & -1 & 1 \\ \hline \text{Case 6} & -1 & 1 & 1 \\ \hline \text{Case 7} & 1 & -1 & 1 \\ \hline \text{Case 8} & 1 & 1 & -1 \end{array}\]

Observe that,because of the symmetry of the expression,Cases 3,4,5 and Cases 6,7,8 are equivalent.Therefore,we only need to check Case 1,2,3 and 6.Simply evaluate the cases to get that \(abc(a^3-b^3)(b^3-c^3)(c^3-a^3)\equiv 0\pmod{7}\;\forall \;a,b,c\in \mathbb{Z}\) Abdur Rehman Zahid · 1 month, 2 weeks ago

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@Abdur Rehman Zahid Nice solution bro..........upvoted Abhisek Mohanty · 1 month, 1 week ago

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vmc questions Piyush Kumar Behera · 1 month, 2 weeks ago

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Can anyone recommend me some good books for INMO and and other maths olympiad?????????? Abhisek Mohanty · 1 month, 2 weeks ago

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